A parallel-plate capacitor is charged to a potential difference and then disconnected…
PHYSICS · Single correct · CBSE · Physics · Electrostatic Potential and Capacitance · Electrostatic Potential and Capacitance
Problem
A parallel-plate capacitor is charged to a potential difference ΔV and then disconnected from the battery. A dielectric slab with dielectric constant κ=3 is fully inserted between the plates. Which statement correctly describes how the energy stored in the capacitor changes once the slab is in place?
(a) The stored energy increases to three times its original value because the capacitance triples while charge remains constant.
(b) The stored energy remains the same because the charge and the potential difference are unchanged.
(c) The stored energy decreases to one-third of its original value because the capacitance triples while charge remains constant.
(d) The stored energy decreases by a factor of 3 because the electric field magnitude is reduced by that factor.